Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > work1, work2;
int n, a[10005], b[10005];
inline void gauss1() {
  int now = 1;
  for (int bit = 30; bit >= 0; bit--) {
    for (int i = now; i <= n; i++)
      if ((b[i] >> bit) & 1) {
        if (i != now) {
          work1.push_back(make_pair(now, i));
          b[now] ^= b[i];
        }
        break;
      }
    if (!((b[now] >> bit) & 1)) continue;
    for (int i = 1; i <= n; i++)
      if (i != now && ((b[i] >> bit) & 1)) {
        work1.push_back(make_pair(i, now));
        b[i] ^= b[now];
      }
    now++;
  }
}
inline void gauss2() {
  int now = 1;
  for (int bit = 30; bit >= 0; bit--) {
    for (int i = now; i <= n; i++)
      if ((a[i] >> bit) & 1) {
        if (i != now) {
          work2.push_back(make_pair(now, i));
          a[now] ^= a[i];
        }
        break;
      }
    if (!((a[now] >> bit) & 1)) continue;
    for (int i = 1; i <= n; i++)
      if (i != now && (((a[i] >> bit) & 1) != ((b[i] >> bit) & 1))) {
        work2.push_back(make_pair(i, now));
        a[i] ^= a[now];
      }
    if (!((b[now] >> bit) & 1)) {
      work2.push_back(make_pair(now, now));
      a[now] = 0;
      continue;
    }
    now++;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  gauss1();
  gauss2();
  for (int i = 1; i <= n; i++)
    if (a[i] != b[i]) {
      puts("-1");
      return 0;
    }
  printf("%d\n", work1.size() + work2.size());
  for (int i = 0; i < work2.size(); i++)
    printf("%d %d\n", work2[i].first, work2[i].second);
  for (int i = work1.size() - 1; i >= 0; i--)
    printf("%d %d\n", work1[i].first, work1[i].second);
  return 0;
}